BZOJ 1038 半平面交

多么明显的半平面交。

答案一定在山的顶点处或者半平面交区域的顶点处。。

 

 

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  1 #include <iostream>

  2 #include <cstring>

  3 #include <cstdlib>

  4 #include <cstdio>

  5 #include <algorithm>

  6  

  7 #define N 555

  8 #define EPS 1e-7

  9 #define INF 1e12

 10  

 11 using namespace std;

 12  

 13 struct PO

 14 {

 15     double x,y;

 16     void prt() {printf("%lf     %lf\n",x,y);}

 17 }p[N],tp[N],s[N];

 18  

 19 int n,m;

 20  

 21 inline PO operator +(PO a,PO b)

 22 {

 23     a.x+=b.x; a.y+=b.y;

 24     return a;

 25 }

 26  

 27 inline PO operator -(PO a,PO b)

 28 {

 29     a.x-=b.x; a.y-=b.y;

 30     return a;

 31 }

 32  

 33 inline PO operator *(PO a,double k)

 34 {

 35     a.x*=k; a.y*=k;

 36     return a;

 37 }

 38  

 39 inline PO operator /(PO a,double k)

 40 {

 41     a.x/=k; a.y/=k;

 42     return a;

 43 }

 44  

 45 inline int dc(double x)

 46 {

 47     if(x>EPS) return 1;

 48     else if(x<-EPS) return -1;

 49     return 0;

 50 }

 51  

 52 inline double cross(const PO &a,const PO &b,const PO &c)

 53 {

 54     return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);

 55 }

 56  

 57 inline double dot(const PO &a,const PO &b,const PO &c)

 58 {

 59     return (b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y);

 60 }

 61  

 62 inline PO getpoint(const PO &a,const PO &b,const PO &c,const PO &d)

 63 {

 64     double k1=cross(a,d,c),k2=cross(b,c,d);

 65     return a+(b-a)*k1/(k1+k2);

 66 }

 67  

 68 inline bool onseg(const PO &a,const PO &b,const PO &c)

 69 {

 70     if(dc(dot(a,b,c))<=0) return true;

 71     return false;

 72 }

 73  

 74 inline void read()

 75 {

 76     scanf("%d",&n);

 77     for(int i=1;i<=n;i++) scanf("%lf",&p[i].x);

 78     for(int i=1;i<=n;i++) scanf("%lf",&p[i].y);

 79 }

 80  

 81 inline void getcut()

 82 {

 83     tp[1].x=tp[5].x=-INF; tp[1].y=tp[5].y=-INF;

 84     tp[2].x=INF,tp[2].y=-INF;

 85     tp[3].x=INF,tp[3].y=INF;

 86     tp[4].x=-INF,tp[4].y=INF;

 87     int cp=4,tc;

 88     for(int i=1;i<n;i++)

 89     {

 90         tc=0;

 91         for(int j=1;j<=cp;j++)

 92         {

 93             if(dc(cross(p[i],p[i+1],tp[j]))>=0) s[++tc]=tp[j];

 94             if(dc(cross(p[i],p[i+1],tp[j])*cross(p[i],p[i+1],tp[j+1]))<0)

 95                 s[++tc]=getpoint(p[i],p[i+1],tp[j],tp[j+1]);

 96         }

 97         s[tc+1]=s[1];

 98         for(int j=1;j<=tc+1;j++) tp[j]=s[j];

 99         cp=tc;

100     }

101     m=cp;

102 }

103  

104 inline void go()

105 {

106     getcut();

107     double ans=INF;

108     PO s1,t1,jd;

109     for(int i=1;i<=n;i++)

110     {

111         s1=p[i]; t1.x=s1.x,t1.y=520.1314;

112         for(int j=1;j<=m;j++)

113         {

114             jd=getpoint(s1,t1,s[j],s[j+1]);

115             if(onseg(jd,s[j],s[j+1])) ans=min(ans,jd.y-p[i].y);

116         }

117     }

118     for(int i=1;i<=m;i++)

119     {

120         s1=s[i]; t1.x=s[i].x; t1.y=520.1314;

121         for(int j=1;j<n;j++)

122         {

123             jd=getpoint(s1,t1,p[j],p[j+1]);

124             if(onseg(jd,p[j],p[j+1])) ans=min(ans,s[i].y-jd.y);

125         }

126     }

127     printf("%.3lf\n",ans);

128 }

129  

130 int main()

131 {

132     read(),go();

133     return 0;

134 }